can anyone gave me some idea of how to solve $a_{n+2}-2\cos(\theta)a_{n+1}+a_n=0$ ? I am not sure how to perform a calculation to this three term recurrence-- I know I can solve a two term recurrence but a characteristic polynomial but I really have no idea how to do this-- some help will be appreciated. Thanks!
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3The characteristic polynomial is $z^2-(2\cos\theta)z+1$. Its zeros are $e^{i\theta}$ and $e^{-i\theta}$. – GEdgar Mar 28 '18 at 01:24
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thank you! just discovered that this is indeed characteristic polynomial – M. Chen Mar 28 '18 at 01:29
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As mentioned by GEdgar, the characteristic polynomial is $$ z^2 - 2 \cos(\theta)z + 1 \text{.} $$ It's worth pointing out that "$\cos \theta$" is just some constant, so you could replace it with some parameter and ignore its trigonometry-ness. The characteristic polynomial factors as $$ (z - (\cos \theta + \mathrm{i} \sin \theta))(z - (\cos \theta - \mathrm{i} \sin \theta)) \text{,} $$ so the general solution to the recurrence is $$ C_1(\cos \theta + \mathrm{i} \sin \theta)^n + C_2(\cos \theta - \mathrm{i} \sin \theta)^n \text{.} $$
We use the fact(s): $$ 1 = \cos^2(\theta) + \sin^2(\theta) = (\cos \theta + \mathrm{i} \sin \theta)(\cos \theta - \mathrm{i} \sin \theta) \text{.} $$
Eric Towers
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also is there a way to compute c1 and c2 if we do not know what a0 is? – M. Chen Mar 28 '18 at 01:47
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@M.Chen : We have the general solution. Since you have a three term recurrence, we need to know (usually, the first) two terms to get a particular solution. From any two $a_i, a_j$, you can find $C_1$ and $C_2$ from $$C_1(\cos \theta + \mathrm{i} \sin \theta)^i + C_2(\cos \theta - \mathrm{i} \sin \theta)^i = a_i$$ and the analogous equation for $a_j$. (This is straightforward algebra since these are linear equations.) – Eric Towers Mar 28 '18 at 02:00